A solution contains 0-01mol L -1 of each Pb 2- and Zn 2- ions- The solution is saturated with H 2 S when - S 2- - is 1-0- 10 -14 mol L -1- Predict which one of the two ions will be precipitated from the solution- K sp - PbS-2-4- 10 -27 and K sp - ZnS-1-0- 10 -21

PbS ⇌ Pb^2++S^2 #160; #160; #160; #160; #160; ZnS ⇌ Zn^2++S^2 Now, [Pb^2+] [Zn^2+]= 0.01 mol L^ 1 [S^2 ]=1.0 × 10^ 14molL^ 1 ...

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Byju's Answer

Standard XII

Chemistry

Common Ion Effect

A solution co...

Question

A solution contains $0.01 m o l L^{- 1}$ of each ${P b}^{2 +}$ and ${Z n}^{2 +}$ ions. The solution is saturated with $H_{2} S$ when $[S^{2 -}]$ is $1.0 \times 10^{- 14} m o l L^{- 1}$ . Predict which one of the two ions will be precipitated from the solution?
( $K_{s p} : P b S = 2.4 \times 10^{- 27}$ and $K_{s p} : Z n S = 1.0 \times 10^{- 21}$ )

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Solution

$P b S ⇌ P b^{2 +} + S^{2 -}$ $Z n S ⇌ Z n^{2 +} + S^{2 -}$
Now, $[P b^{2 +}] - [Z n^{2 +}] = 0.01 m o l L^{- 1}$
$[S^{2 -}] = 1.0 \times 10^{- 14} m o l L^{- 1}$
Now, $[P b^{2 +}] [S^{2 -}] = 0.01 \times 1 \times 10^{- 14} = 1 \times 10^{- 16}$
Now, $[P b^{2 +}] [S^{2 -}] > K_{s p} (P b S)$
$∴ P b S$ will be precipitated.
Again, $[Z n^{2 +}] [S^{2 -}] > K_{s p} (Z n S)$
$∴ Z n S$ will also be precipitated.

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A solution contains 0.01mol L−1 of each Pb2+ and Zn2+ ions. The solution is saturated with H2S when [S2−] is 1.0×10−14molL−1. Predict which one of the two ions will be precipitated from the solution? (Ksp: PbS=2.4×10−27 and Ksp: ZnS=1.0×10−21)

PbS⇌Pb2++S2− ZnS⇌Zn2++S2−Now, [Pb2+]−[Zn2+]=0.01molL−1[S2−]=1.0×10−14molL−1Now, [Pb2+][S2−]=0.01×1×10−14=1×10−16Now, [Pb2+][S2−]>Ksp(PbS)∴PbS will be precipitated.Again, [Zn2+][S2−]>Ksp(ZnS)∴ZnS will also be precipitated.